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Metamath Proof Explorer

Theorem eqrd

Description: Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017) (Proof shortened by BJ, 1-Dec-2021)

	Ref	Expression
Hypotheses	eqrd.0	⊢ Ⅎ 𝑥 𝜑
	eqrd.1	⊢ Ⅎ 𝑥 𝐴
	eqrd.2	⊢ Ⅎ 𝑥 𝐵
	eqrd.3	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
Assertion	eqrd	⊢ ( 𝜑 → 𝐴 = 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		eqrd.0	⊢ Ⅎ 𝑥 𝜑
2		eqrd.1	⊢ Ⅎ 𝑥 𝐴
3		eqrd.2	⊢ Ⅎ 𝑥 𝐵
4		eqrd.3	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
5	1 4	alrimi	⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
6	2 3	cleqf	⊢ ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
7	5 6	sylibr	⊢ ( 𝜑 → 𝐴 = 𝐵 )